How to find the tension in a guitar string?

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SUMMARY

The tension in a guitar string can be calculated using the formula f = (1/2l)√(T/μ), where f is the fundamental frequency (247.0 Hz), l is the length of the string (63.5 cm), T is the tension, and μ is the mass per unit length. The B-string, made of steel with a density of 7800 kg/m³ and a diameter of 0.406 mm, requires understanding of Young's modulus and mass distribution to determine tension accurately. The relationship between frequency and tension is crucial, as higher tension results in a higher pitch. This discussion clarifies the importance of tension in string instruments and provides a mathematical approach to solving the problem.

PREREQUISITES
  • Understanding of fundamental frequency in string instruments
  • Knowledge of Young's modulus and its application
  • Familiarity with mass per unit length (μ) calculations
  • Basic principles of harmonic motion and differential equations
NEXT STEPS
  • Study the derivation of the wave equation for vibrating strings
  • Learn about Young's modulus and its significance in material science
  • Explore the relationship between tension and pitch in string instruments
  • Investigate the effects of string diameter and density on tension
USEFUL FOR

Musicians, physics students, and anyone interested in the mechanics of string instruments will benefit from this discussion, particularly those looking to understand the relationship between tension and frequency in guitar strings.

JustinLiang
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Homework Statement


The B-string of a guitar is made of steel (density 7800 kg/m^3), is 63.5 cm long, and has diameter 0.406mm. The fundamental frequency is f = 247.0 Hz. Find the string tension.

Homework Equations


F/A = YΔL/L

The Attempt at a Solution


So I know we have the A, the Young's modulus, and the length. But there is no change in length! So there should be no tension/stress. I have no clue where to start, need some hints!
Also how does frequency have anything to do with the tension? Is it just there to throw you off?

Thanks.
 
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Frequency has a lot to do with tension. The length from the nut to the bridge does not change, but the tension does. You have to adjust the tension in the string using the tuning machine. Higher tension = higher pitch.
 
You should be supplied with the formula..

f = \frac{1}{2l}\sqrt{\frac{T}{\mu}}

f = frequency
l = length
T = tension
\mu = mass per unit length
 
Darth Frodo said:
You should be supplied with the formula..

f = \frac{1}{2l}\sqrt{\frac{T}{\mu}}

f = frequency
l = length
T = tension
\mu = mass per unit length

Wow, never seen that before, I think our prof assigned the wrong question. I talked to a lot of my friends and they have no clue what to do with this question either.

Thanks though, now I know :P
 
for simple harmonic motion, F=k x, where x is the displacement from equilibrium...

Into Newton's Second Law, m\frac{d^2x}{dt^2}=kx, you get a differential equation whose solution x=A\cos(\omega t+\theta_0) tells you what
the angular frequency \omega is (and what it depends and doesn't depend on).

Can you use "relevant equation" here?
 

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